Problem: Divide the following complex numbers. $ \dfrac{5+45i}{5-5i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5+5i}$ $ \dfrac{5+45i}{5-5i} = \dfrac{5+45i}{5-5i} \cdot \dfrac{{5+5i}}{{5+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(5+45i) \cdot (5+5i)} {(5-5i) \cdot (5+5i)} = \dfrac{(5+45i) \cdot (5+5i)} {5^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(5+45i) \cdot (5+5i)} {(5)^2 - (-5i)^2} = $ $ \dfrac{(5+45i) \cdot (5+5i)} {25 + 25} = $ $ \dfrac{(5+45i) \cdot (5+5i)} {50} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({5+45i}) \cdot ({5+5i})} {50} = $ $ \dfrac{{5} \cdot {5} + {45} \cdot {5 i} + {5} \cdot {5 i} + {45} \cdot {5 i^2}} {50} $ Evaluate each product of two numbers. $ \dfrac{25 + 225i + 25i + 225 i^2} {50} $ Finally, simplify the fraction. $ \dfrac{25 + 225i + 25i - 225} {50} = \dfrac{-200 + 250i} {50} = -4+5i $